Convergence in distribution of the log-Gamma distribution

Question

Suppose $X$ has density $f(x)=\exp(kx-e^x)/\Gamma(k)$, $x>0$, for some parameter $k>0$. Then the moment-generating function of $X$ has the form $$ M_X(\theta)=\frac{\Gamma(\theta+k)}{\Gamma(k)}. $$ I want to show that $$ \lim_{k\to\infty}M_{X^*}(\theta)=\exp(\theta^2/2), $$ where $X^*=\sqrt{k}(X-\log k)$ using Stirling's formula: $$ \log \Gamma(k)=-k+\left(k-\tfrac12\right)\log k+\log\sqrt{2\pi}+O(k^{-1}). $$ Now $$ M_{X^*}(\theta)={\rm E}[\exp(\theta\sqrt{k}(X-\log k))]=\exp(-\theta\sqrt{k}\log k)M_X(\theta\sqrt{k}) $$ and hence $$ \log M_{X^*}(\theta)=-\theta\sqrt{k}\log k+\log \Gamma(\theta\sqrt{k}+k)-\log \Gamma(k). $$ Using Stirling's formula, I obtain $$ \begin{align} \log M_{X^*}(\theta)=&-\theta\sqrt{k}\log k-(\theta\sqrt{k}+k)+(\theta\sqrt{k}+k-\tfrac12)\log(\theta\sqrt{k}+k)+\log\sqrt{2\pi}+O(k^{-1})\\ &+k-\left(k-\tfrac12\right)\log k-\log\sqrt{2\pi}-O(k^{-1})\\ \end{align} $$ but nothing really seems to cancel out. How do I proceed from here? Thanks.

Answer 1

Use the decomposition $\log(\theta\sqrt{k}+k)=\log(k)+\log(1+\theta/\sqrt{k})$ and the expansion $\log(1+\theta/\sqrt{k})=\theta/\sqrt{k}-\theta^2/(2k)+o(1/k)$ then watch the simplifications happening.

As a way of confirmation, one can note that the density $f_k$ of $X^*$ is such that $f_k(x)=c_k\mathrm e^{-u_k(x)}$ with $$ c_k=\frac{\mathrm e^{-k}k^{k-1/2}}{\Gamma(k)},\qquad u_k(x)=k(\mathrm e^{x/\sqrt{k}}-1)-\sqrt{k}x, $$ and that $c_k\to\frac1{\sqrt{2\pi}}$ and $u_k(x)\to\frac12x^2$ when $k\to\infty$.